If random variable $X$ has moments of all orders, does this imply that the moment generating function of $X$ exist?

Question

Suppose $EX^n<\infty$ for any $n$, does this imply that $M_X(t)=Ee^{tX}<\infty$ for all $t$ in some neighborhood of zero, that is, the moment generating function of $X$ exist? If this is not true, a counterexample would be very helpful. Thanks!

Answer 1

Let $f(x)=ce^{-\sqrt n}$ for $n <x<n+1$, $n=1,2,\ldots$ where $c$ is chosen such that $\int _0^{\infty} f(x)dx=1$. I will let you show that this is a density function of a r.v. which has finite moments of all orders but $\int e^{tx} f(x)dx=\infty$ for every $t >0$. [It helps to write $tn-\sqrt n=\sqrt n (t\sqrt n -1)$. Get a lower bound for this on $(n,n+1)$].

As pointed out by Adayah below one can just take $f(x)=ce^{-\sqrt x}$ for all $x >0$.

Answer 2

The series $$ \sum_{n=0}^{\infty}\frac{t^nE(X^n)}{n!} $$ converges for all $t$ in a nbd of zero if and only the radius of convergence is positive, i.e. if and only if $$ \overline{\lim}_{n\to\infty}\left(\frac{|E(X^n)|}{n!}\right)^{1/n}<\infty $$

Since $(n!)^{1/n}\approx (n/e)$ as $n\to\infty$, this imposes a growth condition on the sequence $\frac{|E(X^n)|^{1/n}}{n}$. Thus, for example, if $X$ is bounded, then the MGF will exist everywhere, (because the sequence $|E(X^n)|^{1/n}$ tends to the essential supremum of $X$ as $n\to\infty$) but in the general case one could have a MGF that converges nowhere.